Question

What is the pH of a 0.5 M solution of NaHCO3? Kb = 2.4x10-8, hint this is a base!

Answer 1

HCO3- + H2O ---------------------> H2CO3 + OH-

0.5                                                      0           0   ----------------> initial

0.5-x                                                  x              x -----------------> equilibrium

Kb = [H2CO3][OH-]/[HCO3-]

Kb = x^2 / 0.5 - x

2.4 x 10^-8 = x^2 / 0.5 -x

x^2 + 2.4 x 10^-8 x - 1.2 x 10^-8 = 0

x = 1.095 x 10^-4

[OH-] = x = 1.095 x 10^-4 M

pOH = -log [OH-]

pOH = -log (1.095 x 10^-4)

pOH = 3.96

pH + pOH = 14

pH = 10.04